A remote sensing satellite of the Earth revolves in a circular orbit at a height of 250 km above the Earth's surface. What is the (i) Orbital speed and (ii) period of revolution of the satellite. Radius of the Earth, `R=6.38xx10^(6)m`, and acceleration due to gravity on the surface of the Earth, `g=9.8ms^(-2)`

Explanation

(i) Here `g=9.8ms^(-2),R=6.38xx10^(6)m`
h = 2,50,000 m
`R+h=6.38xx10^(6)+2,50,000=6.63xx10^(6)m`
The orbial speed is given by
`V_(0)=sqrt((gR^2)/(R+h))=sqrt((9.8xx(6.38xx10^(6))^(2))/(6.63xx10^(6)))`
`=7.76xx10^(3)ms^(-1)=7.76kms^(-1)`
(ii) The period of revolution of the satellite will be
`T=(2pi(R+h))/(V_0)=(2xx22xx6.63xx10^(6))/(7xx7.76xx10^(3))=5370s`